Here, we use the fact that the inverse of a product of elements is the product of their inverses, in reverse order (i.e. the inverse map is involutive). Thus, we have:

Note that for the proof to go through we need to use the fact that and are left and right inverses of each other.

NOTE: It actually suffices to prove only the first of these three things, because to test whether a map between groups is a homomorphism of groups, it suffices to check whether it sends products to products. However, when working in somewhat greater generality than groups, it becomes important to check the other conditions, and they're explained here for illustrative purposes.

Thus, every is a homomorphism. It remains to show that this homomorphism is injective and surjective. Injectivity is clear, because . For surjectivity, note that given any , setting:

yields . (This also becomes clearer from the fact that the map is itself a homomorphism of groups).